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The enthalpy of neutralisation of HCl and NaOH is -57 kJ mol-1. The heat evolved at constant pressure (in kJ) when 0.5 mole of H2SO4 react with 0.75 mole of NaOH is equal to
- a)57 ×
- b)57 × 0.5
- c)57
- d)57 × 0.25
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The enthalpy of neutralisation of HCl and NaOH is -57 kJ mol-1. The he...
-57 kJ of heat evolved when 1 mole of NaOH reacted with an acid.
We have 1 mole of H+ and 0.75 moles of OH-. So OH- iis limiting reagent. Or only 0.75 moles of OH- will be used. So for 1 mole of OH-, we have -57 kJ heat released. Therefore for 0.75 moles, we have ¾× -57 kJ heat released.
We have 1 mole of H+ and 0.75 moles of OH-. So OH- iis limiting reagent. Or only 0.75 moles of OH- will be used. So for 1 mole of OH-, we have -57 kJ heat released. Therefore for 0.75 moles, we have ¾× -57 kJ heat released.
Most Upvoted Answer
The enthalpy of neutralisation of HCl and NaOH is -57 kJ mol-1. The he...
-57 kJ of heat evolved when 1 mole of NaOH reacted with an acid.
We have 1 mole of H+ and 0.75 moles of OH-. So OH- iis limiting reagent. Or only 0.75 moles of OH- will be used. So for 1 mole of OH-, we have -57 kJ heat released. Therefore for 0.75 moles, we have ¾× -57 kJ heat released.
We have 1 mole of H+ and 0.75 moles of OH-. So OH- iis limiting reagent. Or only 0.75 moles of OH- will be used. So for 1 mole of OH-, we have -57 kJ heat released. Therefore for 0.75 moles, we have ¾× -57 kJ heat released.
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Community Answer
The enthalpy of neutralisation of HCl and NaOH is -57 kJ mol-1. The he...
You have calculated -57 Kj of heat evolved when I mole of NaOH reacted with an acid.
Obviously heat evolved will be reduced with the reduction in number of moles....therefore
since 0.75 is 3/4 the answer is A.
Hope this answer helps
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The enthalpy of neutralisation of HCl and NaOH is -57 kJ mol-1. The heat evolved at constant pressure (in kJ) when 0.5 mole of H2SO4react with 0.75 mole of NaOH is equal toa)57 ×b)57 × 0.5c)57d)57 × 0.25Correct answer is option 'A'. Can you explain this answer?
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The enthalpy of neutralisation of HCl and NaOH is -57 kJ mol-1. The heat evolved at constant pressure (in kJ) when 0.5 mole of H2SO4react with 0.75 mole of NaOH is equal toa)57 ×b)57 × 0.5c)57d)57 × 0.25Correct answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The enthalpy of neutralisation of HCl and NaOH is -57 kJ mol-1. The heat evolved at constant pressure (in kJ) when 0.5 mole of H2SO4react with 0.75 mole of NaOH is equal toa)57 ×b)57 × 0.5c)57d)57 × 0.25Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The enthalpy of neutralisation of HCl and NaOH is -57 kJ mol-1. The heat evolved at constant pressure (in kJ) when 0.5 mole of H2SO4react with 0.75 mole of NaOH is equal toa)57 ×b)57 × 0.5c)57d)57 × 0.25Correct answer is option 'A'. Can you explain this answer?.
The enthalpy of neutralisation of HCl and NaOH is -57 kJ mol-1. The heat evolved at constant pressure (in kJ) when 0.5 mole of H2SO4react with 0.75 mole of NaOH is equal toa)57 ×b)57 × 0.5c)57d)57 × 0.25Correct answer is option 'A'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The enthalpy of neutralisation of HCl and NaOH is -57 kJ mol-1. The heat evolved at constant pressure (in kJ) when 0.5 mole of H2SO4react with 0.75 mole of NaOH is equal toa)57 ×b)57 × 0.5c)57d)57 × 0.25Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The enthalpy of neutralisation of HCl and NaOH is -57 kJ mol-1. The heat evolved at constant pressure (in kJ) when 0.5 mole of H2SO4react with 0.75 mole of NaOH is equal toa)57 ×b)57 × 0.5c)57d)57 × 0.25Correct answer is option 'A'. Can you explain this answer?.
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